Difference between revisions of "2003 AIME I Problems/Problem 1"

Revision as of 09:17, 20 April 2008

Given that

$\frac{((3!)!)!}{3!} = k \cdot n!,$

where $k$ and $n$ are positive integers and $n$ is as large as possible, find $k + n.$

We use the definition of a factorial to get

$\frac{((3!)!)!}{3!} = \frac{(6!)!}{3!} = \frac{720!}{3!} = \frac{720!}{6} = \frac{720 \cdot 719!}{6} = 120 \cdot 719! = k \cdot n!$

We certainly can't make $n$ any larger if $k$ is going to stay an integer, so the answer is $k + n = 120 + 719 = 839$ .

@@ Line 14: / Line 14: @@
 == See also ==
-* [[2003 AIME I Problems/Problem 2 | Next problem]]
+{{AIME box|year=2002|n=I|before=First Question|num-a=2}}
-* [[2003 AIME I Problems]]
 [[Category:Introductory Number Theory Problems]]

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